Solve for $z$, which satisfy $arg(z-3-2i) = frac{pi}{6}$ and $arg(z-3-4i) = frac{2pi}{3}$.
up vote
2
down vote
favorite
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
|
show 1 more comment
up vote
2
down vote
favorite
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07
|
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
solve for $z$, which satisfy $displaystyle arg(z-3-2i) = frac{pi
}{6}$ and $displaystylearg(z-3-4i) = frac{2pi}{3}$.
So I'm first assuming $z=x+iy$, then putting it in the place of $z$ and putting real parts together and imaginary parts together.
Then I'm using $tantheta = frac{text{imaginary part}}{text{real part}}$
$z-3-4i = (x-3)+(y-2)i$
Then,
$tan30° = frac{y-2}{x-3}$
$frac{1}{sqrt3} = frac{y-2}{x-3}$
I'm processing like this and my answer comes as $y=5/2$
And $x= 3+frac{5sqrt3}{2}-2sqrt3$
But the answer key says there's no such $z$ which satisfies this equation.
Is this the wrong way to solve this question, is my answer wrong or the answer key's?
complex-analysis complex-numbers
complex-analysis complex-numbers
edited Nov 22 at 4:37
Kemono Chen
2,050435
2,050435
asked Nov 22 at 3:53
Kaustuv Sawarn
465
465
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07
|
show 1 more comment
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
1
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008723%2fsolve-for-z-which-satisfy-argz-3-2i-frac-pi6-and-argz-3-4i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
add a comment |
up vote
2
down vote
accepted
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
$$arg(z-3-2i)=frac{pi}{6}$$ is a line which originate from $(3,2)$ and making an angle of $30^circ$ with positive $x$ axis.
And $$arg(z-3-4i)=frac{2pi}{3}$$ is a line which originate from $(3,4)$ and making an angle of $120^circ$ with positive $x$ axis.
Now drawing These line in $x-y$ Coordinate axis.
You will get no point of Intersection.
So no $z$ which satisfy above these two equations.
answered Nov 22 at 4:03
Durgesh Tiwari
5,2862630
5,2862630
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
add a comment |
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Okay, but is my approach wrong? If it is, where is it wrong?
– Kaustuv Sawarn
Nov 22 at 4:05
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
Actually you have take above line as infinite ray.
– Durgesh Tiwari
Nov 22 at 4:06
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3008723%2fsolve-for-z-which-satisfy-argz-3-2i-frac-pi6-and-argz-3-4i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Why are you equating the arguments of the $Arg$'s? Or should that $2$ be a $4$?
– Michael Burr
Nov 22 at 4:00
1
It may be useful to draw a diagram.
– Kemono Chen
Nov 22 at 4:04
@MichaelBurr i didn't get you. I'm not equating the arguments.
– Kaustuv Sawarn
Nov 22 at 4:06
@KemonoChen but I get the answer this way, i want to know what am I doing wrong?
– Kaustuv Sawarn
Nov 22 at 4:07
Your approach is wrong .
– Akash Roy
Nov 22 at 4:07